• Apr 25th 2012, 10:36 AM
Kibbygirl
Factor completely:

1. 5k3- 40

5(k-2)(k2 + 2k + 4)

2. t2 - 81/169

(t + 9/13)(t - 9/13)

3. 12b2+ 36b + 27

(2b+3)2

4. b4- 16

(b2 +4)(b+2)(b-2)

82. (2k-7)h2 - 4 (2k-7)h-45(2k-7)

(2k-7)(h+5)(h-9)

Thank you :)
• Apr 25th 2012, 10:46 AM
biffboy
Correct except that you have lost a factor of 3 in no.3 Should be 3(2b+3)^2
• Apr 25th 2012, 10:48 AM
HallsofIvy
Quote:

Originally Posted by Kibbygirl
Factor completely:

1. 5k3- 40

5(k-2)(k2 + 2k + 4)

Yes, that is correct.

Quote:

2. t2 - 81/169

(t + 9/13)(t - 9/13)
Yes, that is correct.

Quote:

3. 12b2+ 36b + 27

(2b+3)2
No, that is not correct. What you have written is the same as 2(2b+ 3)= 4b+ 6 (multiplication is commutative). What I think you meant was
(2b+ 3)2 but that is still incorrect. (2b+ 3)2= (2b)2+ 2(2b)(3)+ 32= 4b2+ 12b+ 9. Did you leave out a "3" that you factored out first? 12b2+36b+ 27= 3(4b2+ 12b2+ 9)= 3(2b+ 3)2.

Quote:

4. b4- 16

(b2 +4)(b+2)(b-2)
Yes, this is correct.

Quote:

82. (2k-7)h2 - 4 (2k-7)h-45(2k-7)

(2k-7)(h+5)(h-9)
Yes, that is correct.

Quote:

Thank you :)